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Question Number 70312 by Shamim last updated on 03/Oct/19

$$\mathrm{If},\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}+\frac{1}{{\mathrm{c}}^2}=\frac{1}{\mathrm{ab}}+\frac{1}{\mathrm{bc}}+\frac{1}{\mathrm{ca}}\mathrm{then}\mathrm{prove}$$$$\mathrm{that},\mathrm{a}=\mathrm{b}=\mathrm{c}.$$

Commented by MJS last updated on 03/Oct/19

$$\mathrm{this}\mathrm{is}\mathrm{true}\mathrm{for}a,b,c\in \mathbb{R}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{for}a,b,c\in \mathbb{C}$$$$b=a(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})\wedge c=a(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})$$$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=a\times \left(\begin{array}{c}1\\ -\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\\ -\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i}\end{array}\right)\mathrm{is}\mathrm{a}\mathrm{solution}$$$$\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}a\\ b\\ \frac{ab(a+b)\pm \sqrt{3}ab(a-b)\mathrm{i}}{2(a^2-ab+b^2)}\end{array}\right)\mathrm{is}\mathrm{a}\mathrm{solution}$$

Answered by $@ty@m123 last updated on 03/Oct/19

$$Let\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z$$$$\Rightarrow x^2+y^2+z^2=xy+yz+zx$$$$\Rightarrow x^2+y^2+z^2-(xy+yz+zx)=0$$$$\Rightarrow 2\{x^2+y^2+z^2-(xy+yz+zx)\}=0$$$$\Rightarrow {(x-y)}^2+{(y-z)}^2+{(z-x)}^2=0$$$$\Rightarrow x-y=0,y-z=0,z-x=0$$$$\Rightarrow x=y=z$$$$\Rightarrow a=b=c$$

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